Question

The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly
(radius of the nucleus $=1.1$ fermi) $\left[\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right]$

• A. 0.65 MeV
• B. 2.09 MeV
• C. 3.31 MeV
• D. 0.92 MeV

Answer 1

Answer: (A)

Potential energy due to two charges $=\frac{k q_{1} q_{2}}{d}$.
For hydrogen atom, $q_{1}=q_{2}=1.6 \times 10^{-19} C$
$U =\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{2.2 \times 10^{-15}} [\therefore d=2s]$
$= 10.47 \times 10^{-14} J =\frac{10.47 \times 10^{-4}}{1.6 \times 10^{-19}}\, eV$
$=6.5 \times 10^{5} eV =0.65\, MeV$